Download Unit 5 DNA/RNA/PROTEIN SYNTHESIS

BIOLOGY I HONORS
CAPACITY MATRIX
UNIT V: DNA
Mrs. Jamie Edmonds, M.Ed., NBCT
Mayo High School for Math, Science & Technology
[email protected]
398-2681
Standard B-4:
The student will demonstrate an
understanding of the molecular
basis of heredity.
AIM
Heard
of it
INDICATORS
Know it
Can
teach it
Text
References
B-4.1 Compare DNA and RNA in terms of
structure, nucleotides, and base pairs.
B-4.2 Summarize the relationship among
DNA, genes, and chromosomes.
B-4.3 Explain how DNA functions as the code
of life and the blueprint for proteins.
B-4.4 Summarize the basic processes
involved in protein synthesis (including
transcription and translation).
B-4.8 Compare the consequences of
mutations in body cells with those in
gametes.
B-4.1 Compare DNA and RNA in terms of structure, nucleotides, and base pairs.
It is essential for students to understand that nucleic acids are organic molecules that serve as the blueprint for
proteins and, through the action of proteins, for all cellular activity.
 There are two types of nucleic acids.
o Deoxyribonucleic acid (DNA)
o Ribonucleic acid (RNA)
 DNA and RNA are composed of small units called nucleotides. The nucleotides that compose nucleic acids have
three parts:
o A nitrogenous base
 Cytosine (C)
 Guanine (G)
 Adenine (A)
 Thymine (T) (DNA only)
 Uracil (U) (RNA only)
o A simple (pentose) sugar
 Deoxyribose (DNA only)
 Ribose (RNA only)
o A phosphate group
The basic structure of the two molecules is different.
 DNA consists of two single chains which spiral around an imaginary axis to form a double helix with nitrogenous
bases from each strand of DNA chemically bonded through the axis of the helix.
o When the nitrogenous bases of two strands of DNA chemically bond through the center of the helix, each
base can bond to only one type of base. Bases that bond are called complementary bases.
 Guanine (G) will only bond with Cytosine (C).
 Thymine (T) will only bond with Adenine (A).

RNA consists of a single chain of nucleotides with nitrogenous bases exposed along the side.
o When the nitrogenous bases of RNA chemically bond to a strand of DNA, each RNA base can bond with
only one type of DNA base. Bases that bond are called complementary bases.
 Guanine (G) will only bond with Cytosine (C).
 Uracil (U) will only bond with Adenine (A).
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It is essential for students to compare the structure of the two types of nucleic acid.
Type of base composing
nucleotides
Type of sugar composing
nucleotides
Molecule structure and shape
DNA
Cytosine (C)
Adenine (A)
Guanine (G)
Thymine(T)
Deoxyribose
RNA
Cytosine(C)
Adenine (A)
Guanine (G)
Uracil (U)
Ribose
Double helix
Single chain
It is not essential for students to understand
 the chemical formula for DNA or RNA;
 the difference between pyrimidine bases and purine bases.
Assessment Guidelines:
The objective of this indicator is to compare DNA and RNA in terms of structure, nucleotides and base pairs; therefore,
the primary focus of assessment should be to detect similarities and differences between structure of DNA and RNA, the
nucleotides that compose DNA and RNA, and the bases that bond to form DNA and RNA.
In addition to compare, assessments may require students to
 recognize the chemical names of the DNA and RNA molecules;

identify the parts of a nucleotide;
 recognize the names of the 5 bases and the two sugars that compose the nucleotides that make up all nucleic
acids;
 interpret an illustration of a nucleotide;
 interpret an illustration of a DNA or an RNA molecule.
B-4.2 Summarize the relationship among DNA, genes, and chromosomes.
It is essential for students to understand that DNA, genes, and chromosomes compose the molecular basis of
heredity.
 A chromosome is a structure in the nucleus of a cell consisting essentially of one long thread of DNA that is tightly
coiled.
 DNA, composed of nucleotides, provides the blueprint for the synthesis of proteins by the arrangement of
nitrogenous bases.
o The code for a particular amino acid (the base unit of proteins) is determined by a sequence of three base
pairs on the DNA molecule.
 A gene is a specific location on a chromosome, consisting of a segment of DNA, that codes for a particular
protein.
o The particular proteins coded by the DNA on the genes determine the characteristics of an organism.
o Each chromosome consists of hundreds of genes determining the many proteins for an individual
organism.
It is not essential for students to understand the history behind the discovery of DNA.
Assessment Guidelines:
The objective of this indicator is to summarize the relationship among DNA, genes, and chromosomes; therefore, the
primary focus of assessment should be to give major points about how DNA, genes and chromosomes are related.
In addition to summarize, assessments may require students to
 recall the basic structure of chromosomes and genes;
 illustrate or interpret an illustration of the relationship of a chromosome, DNA and genes using words or diagrams.
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B-4.3 Explain how DNA functions as the code of life and the blueprint for proteins.
It is essential for students to understand that the DNA, which comprises the organism’s chromosomes, is considered
the “code of life” (genetic code) because it contains the code for each protein that the organism needs.
 The specificity of proteins is determined by the order of the nitrogenous bases found in DNA.
o In order to construct the specific proteins needed for each specific purpose, cells must have a blueprint
that reveals the correct order of amino acids for each protein found in the organism (thousands of
proteins).
o A gene is a segment of DNA that codes for one particular protein.


Each cell in an organism’s body contains a complete set of chromosomes.
o The number of chromosomes varies with the type of organism. For example, humans have 23 pairs of
chromosomes; dogs have 39 pairs; potatoes have 24 pairs.
o One pair of chromosomes in an organism determines the sex (male, female) of the organism; these are
known as sex chromosomes. All other chromosomes are known as autosomal chromosomes, or
autosomes.
o Cells (except for sex cells) contain one pair of each type of chromosome.
 Each pair consists of two chromosomes that have genes for the same proteins.
 One chromosome in each pair was inherited from the male parent and the other from the female
parent. In this way traits of parents are passed to offspring.
 For example, human cells have 46 chromosomes (23 pairs).
Each chromosome consists of thousands of genes. This is because there are so many unique proteins that each
organism needs to produce in order to live and survive.
o Organisms that are closely related may have genes that code for the same proteins that make the
organisms similar. For example, all maple trees have many of the same genes.
o Each individual organism has unique characteristics and those unique characteristics arise because of
the differences in the proteins that the organism produces.
o Organisms that are not closely related share fewer genes than organisms that are more closely related.
For example, red maple trees share more genes with oak trees than with earthworms.
It is essential for students to understand that DNA can function as the code of life for protein synthesis or the process
of DNA replication, which ensures that every new cell has identical DNA.
 DNA replication is carried out by a series of enzymes. The first enzyme unzips the two strands of DNA that
compose the double helix, separating paired bases.
 Each base that is exposed can only bond to its complementary base.
o Adenine (A) can only bond to thymine (T)
o Cytosine (C) can only bond to guanine (G)
 Each of the separated strands serves as a template for the attachment of complementary bases, forming a new
strand, identical to the one from which it was “unzipped”.
 The result is two identical DNA molecules.
It is not essential for students to understand
 the specific chromosome numbers for organisms, except for humans;
 the names of the specific enzymes needed for replication.
Assessment Guidelines:
The objective of this indicator is to explain how DNA functions as the code of life and the blueprint for proteins, therefore,
the primary focus of assessment should be to construct a cause-and-effect model showing how DNA determines the
functional and structural proteins produced in an organism. Assessment should include how the process of DNA
replication ensures that the entire DNA code is present in every cell of an organism.
In addition to explain, assessments may require students to
 summarize the role of DNA as the code of life;
 summarize the process of DNA replication;
 infer why organisms that are similar in structure or function often share many of the same proteins and genes.
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B-4.4 Summarize the basic processes involved in protein synthesis (including transcription and translation).
It is essential for students to understand that when a particular protein is needed, the cell must make the protein
through the process of protein synthesis. DNA molecules (which contain the code) do not leave the nucleus of the cell, but
protein synthesis must occur in the ribosomes which are located outside of the nucleus in the cytoplasm. Therefore, the
code must be carried from the nucleus to the cytoplasm.
Transcription
Transcription is the process by which a portion of the molecule of DNA is copied into a complementary strand of RNA.
Through the process of transcription, the DNA code is transferred out of the nucleus to the ribosomes.
 Through a series of chemical signals, the gene for a specific protein is turned on. An enzyme attaches to the
exact location on the DNA molecule where the gene is found, causing the two strands of DNA to separate at that
location.
 Complementary RNA nucleotide bases bond to the bases on one of the separated DNA strands.
DNA nucleotide bases exposed on
the separated strand
Adenine (A)
Thymine (T)
Cytosine (C)
Guanine (G)



RNA nucleotide
which bonds
Uracil (U)
Adenine (A)
Guanine (G)
Cytosine (C)
Nucleotides of RNA bond together, forming a single-stranded molecule of RNA that peels away from the DNA
strand and the two DNA strands rejoin. This is called messenger RNA (mRNA).
The messenger RNA (mRNA) is formed complementary to one strand of DNA.
The mRNA strand leaves the nucleus and goes through the nuclear membrane into the cytoplasm of the cell.
Translation
Translation is the process of interpreting the genetic message and building the protein and begins when the mRNA
attaches to a ribosome, which contains proteins and ribosomal RNA (rRNA), in the cytoplasm.
 The function of ribosomes is to assemble proteins according to the code that the mRNA brings from the DNA.
 Each three-base nucleotide sequence on the mRNA is called a codon. Each codon specifies a particular amino
acid that will be placed in the chain to build the protein molecule.
o For example, if the DNA sequence was GAC, then the RNA sequence becomes CUG and the amino acid
that is coded is Leucine.
o TEACHER NOTE: mRNA codons for specific amino acids can be found in tables in most textbooks.
o The sequence of mRNA nucleotides determines the order of the amino acids in the protein chain which,
in turn, distinguishes one protein from another in structure and function.
 Another type of RNA, transfer RNA (tRNA), is vital in assembling amino acids into the correct sequence for the
required protein by transferring amino acids to the ribosomes when needed. There are twenty different types of
tRNA molecules, one for each amino acid.
o At one end of each tRNA is an anticodon site, which has the 3-nucleotide bases complementary to the
codon of mRNA.
o The other end of the tRNA molecule has a specific amino acid attached determined by the anticodon.
 The translation process takes place as follows:
o The tRNA with its attached amino acid pairs to the codon of the mRNA attached to a ribosome.
o When a second tRNA with its specific amino acid pairs to the next codon in sequence, the attached
amino acid breaks from the first tRNA and attaches to the amino acid of the second tRNA.
o The ribosome forms a peptide bond between the amino acids, and an amino acid chain begins to form.
o The empty tRNA moves off and picks up another matching amino acid from the cytoplasm in the cell.
o This sequence is repeated until the ribosome reaches a stop codon on the mRNA, which signals the end
of protein synthesis.
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It is not essential for students to
 understand the details of the processes of transcription and translation, other than as described in the essential
information as given above;
 understand the termination of transcription, in terms of alteration of the mRNA ends and RNA splicing;
 understand or recognize the enzymes involved in the process of protein synthesis;
 know the amino acid that each codon represents;
 recall the specific codon sequence for any amino acid or names of specific amino acids.
Assessment Guidelines:
The objective of this indicator is to summarize the processes involved in protein synthesis, therefore, the primary focus of
assessment should be to give major points about the steps in protein synthesis and the roles of each nucleic acid (DNA,
mRNA, and tRNA) in the processes of transcription and translation.
In addition to summarize, assessments may require students to
 illustrate or interpret illustrations of the processes of transcription, translation, and protein synthesis;
 compare the processes of transcription and translation;
 sequence the steps of transcription and translation;
 explain the significance of each step to the overall process of protein synthesis.
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B-4.8 Compare the consequences of mutations in body cells with those in gametes.
It is essential that students understand that a mutation is the alteration of an organism’s DNA. Mutations can range
from a change in one base pair to the insertion or deletion of large segments of DNA. Mutations can result from a
malfunction during the process of meiosis or from exposure to a physical or a chemical agent, a mutagen.
Most mutations are automatically repaired by the organism’s enzymes and therefore have no effect.
However, when the mutation is not repaired, the resulting altered chromosome or gene structure is then passed to all
subsequent daughter cells of the mutant cell, which may have adverse or beneficial effects on the cell, the organism, and
future generations.
 If the mutant cell is a body cell (somatic cell), the daughter cells can be affected by the altered DNA, but the
mutation will not be passed to the offspring of the organism.
o Body cell mutations can contribute to the aging process or the development of many types of cancer.
 If the mutant cell is a gamete (sex cell), the altered DNA will be transmitted to the embryo and may be passed to
subsequent generations. Gamete cell mutations can result in genetic disorders.
o If the mutation affects a single gene, it is known as a gene mutation.
 For example, the genetic basis of sickle-cell disease is the mutation of a single base pair in the
gene that codes for one of the proteins of hemoglobin.
 Other examples of genetic disorders are Tay-Sachs disease, Huntington’s disease, cystic fibrosis,
or albinism.
o If the mutation affects a group of genes or an entire chromosome, it is known as a chromosomal
mutation.
 Nondisjunction results in an abnormal number of chromosomes, usually occurring during meiosis.
 Examples of abnormalities in humans due to nondisjunction of sex chromosomes are
Klinefelter’s syndrome (male) and Turner’s syndrome (female).
 Examples of abnormalities in humans due to nondisjunction of autosomal chromosomes
include Down syndrome.
In some cases mutations are beneficial to organisms. Beneficial mutations are changes that may be useful to organisms
in different or changing environments. These mutations result in phenotypes that are favored by natural selection and
increase in a population.
It is not essential for students to understand
 the exact mechanism of the various mutations;
 the exact characteristics of the nondisjunction mutation abnormalities listed above;
 the mechanism through which somatic mutations can cause various cancers.
Assessment Guidelines:
The objective of this indicator is to compare the consequences of mutations in body cells with those in gametes, therefore,
the primary focus of assessment should be to detect similarities and differences between the mutations that occur in sex
cells to those in somatic cells.
In addition to compare, assessments may require students to
 recall the causes of mutations;
 classify mutations as resulting from sex cell or somatic cell alterations;
 classify mutations as genetic or chromosomal;
 exemplify genetic or chromosomal disorders;
 explain the effect that various mutations have on the cell, the organism, and future generations.
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